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I want to prove that $(\textbf{a}\times\nabla)\cdot\textbf{b}=-\nabla\cdot(\textbf{a}\times\textbf{b})+\textbf{b}\cdot(\nabla\times\textbf{a})$ using the levi civita symbol, but what does $(\textbf{a}\times\nabla)\cdot\textbf{b}$ even mean in general? If $\textbf{a}=(a_1,a_2,a_3)$ is $(\textbf{a}\times\nabla)_1=a_2\partial_3-a_3\partial_2$ (and so on for other components). When you dot this with $\textbf{b} = (b_1,b_2,b_3)$ do you get (for component 1) $(a_2\frac{\partial b_1}{\partial x_3}-a_3\frac{\partial b_1}{\partial x_2})$?

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  • $\begingroup$ Quite impressive mathjax for a "new user"! $\endgroup$
    – amWhy
    Commented Sep 26, 2018 at 17:34
  • $\begingroup$ How can you prove the identity, when you don't even know what $(\textbf{a}\times\nabla)\cdot\textbf{b}$ even means? $\endgroup$
    – amWhy
    Commented Sep 26, 2018 at 17:35
  • $\begingroup$ @amWhy I think 11Elves asked the meaning of the formula (why would you ever use such thing). Is there maybe a physical significance of such an expression? He/she knows to apply the steps to prove it, but it would be just a mechanical process, without any meaning. $\endgroup$
    – Andrei
    Commented Sep 26, 2018 at 17:40
  • $\begingroup$ I want to prove it, but first i need to know what the left side means. Its not a curl, but something strange. Help $\endgroup$
    – 11Elves
    Commented Sep 26, 2018 at 18:43
  • $\begingroup$ @Andrei At least, that identity is involved in "Energy Conservation" in "Classical Elegtromagnetism". However, I guess there's something wrong with your post. See this link. $\endgroup$ Commented Sep 26, 2018 at 19:13

1 Answer 1

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$\mathbf{a}\times \nabla$ is indeed the operator $(a_2 \partial_3-a_3\partial_2, a_3\partial_1 -a_1\partial_3, a_1\partial_2-a_3\partial_2)$. Taking the dot product with $\mathbf{b}$ gives $(\mathbf{a}\times \nabla)\cdot \mathbf{b}=a_2\partial_3 b_1-a_3\partial_2 b_1 + a_3\partial_1 b_2-a_1\partial_3 b_2 + a_1\partial_2 b_3 - a_2\partial_1 b_3$. However, we can also write this as $a_1(\partial_2 b_3-\partial_3 b_2)+a_2(\partial_3b_1-\partial_1b_3)+a_3(\partial_1 b_2-\partial_2 b_1)=\mathbf{a}\cdot (\nabla \times \mathbf{b})$, which leads you to the formula mentioned by Felix Marin above.

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